Let \(n\) be a positive integer. Evaluate \[ I=\frac{1}{2\pi i}\int_C\frac{z^{n-1}}{3z^n-1}\,dz, \] where \(C\) is the counterclockwise unit circle \(|z|=1\).
Solution
1. The poles of\[ R(z)\coloneqq\frac{z^{n-1}}{3z^n-1} \]are the \(n\) roots of\[ z^n=\frac13, \]and all of them lie inside \(C\).
The equation \(3z^n-1=0\) has exactly \(n\) distinct roots. Every such root \(\alpha\) satisfies \[ \abs\alpha=3^{-1/n}<1, \] so every pole lies strictly inside the unit circle.
2. Each pole \(\alpha\) has residue\[ \operatorname{Res}(R;\alpha)=\frac{1}{3n}. \]
The derivative of the denominator is \[ 3n z^{n-1}. \] At a root \(\alpha\) of \(3z^n-1\), one has \(\alpha\ne0\), so this derivative is nonzero and the pole is simple. Therefore \[ \operatorname{Res}(R;\alpha) = \frac{\alpha^{n-1}}{3n\alpha^{n-1}} = \frac1{3n}. \]
3. The requested integral is\[ \boxed{I=\frac13}. \]
By the residue theorem and steps <1>1–<1>2, \[ I = \sum_{3\alpha^n=1}\operatorname{Res}(R;\alpha) = n\cdot\frac1{3n} = \frac13. \]
4. Q.E.D.
Step 3 gives the required value.